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**Q.1. How many zeroes can the polynomial 2x ^{2 }- 3x + 4 have?**

∴ It can have two zeroes at the most.

∴ y = p (x) has 3 zeroes.

p(x) = x

⇒ p(2) = (2)

⇒ 4 - 6 - 4 = - 6

p(x) = x

⇒ p (- 1) = (- 1)

⇒ 1 + 3 - 4 = 0

The zero of a linear polynomial =

∴ The zero of 2x + 3 = -3/2

The zero of a linear polynomial

∴ The zero of the given linear polynomial

∴ The parabola opens upwards.

p (x) = x

∴ a = 2, b = - 8 and c = 6

⇒ Sum of the zeroes = -b/a

⇒ -(-8)/(2) = 4

As, Product of the zeroes = c/a

⇒ Product of the zeroes of 3x

Sol.

Degree of the divisor f(x) = 2

∴ Degree of the quotient q(x) = Degree of p(x) - Degree of f(x) = 5 - 2 = 3

**Sol. **y = p (x) has its zero at a point where y = 0

In the figure, y = 0 at (2, 0).

∴ 2 is the zero of p (x).**Q.12. For what value of k, (- 4) is a zero of the polynomial x ^{2} - x - (2k + 2)?Sol: **Since, (- 4) is the zero of p (x)

⇒ p (- 4) = (- 4)

⇒ 16 + 4 = 2k + 2

⇒ 20 - 2 = 2k

⇒ 18= 2k

⇒ k = 18/2 = 9

Thus, the required value of k = 9.

∵ p (x) = x

p (- 4) = (- 4)

Since - 4 is a zero of p (x),

∴ (- 4)

⇒ 16 + 8 = 7p + 3

⇒ 24 - 3 = 7p

⇒ 7p = 21

⇒ p = 21/7= 3

Thus, the required value of p is 3.

**Q.14. The sum and product of the zeroes of a quadratic polynomial are and - 3 respectively. What is the quadratic polynomial? ****Sol.** For a quadratic polynomial, we have

Sum of zeroes = -1/2

Product of the zeroes = -3

Since general equation of the quadratic polynomial

= x^{2} - [Sum of the roots] x + [Product of the roots]

⇒ **Q.15. The graph of y = f (x) is given in the figure, find the number of zeroes of f (x).****Sol.** From the figure (graph), it is evident that y = f (x) intersects the x-axis at two distinct points A and B.

∴ Number of zeroes of f (x) = 2.**Q.16. In the adjoining figure the graph of a polynomial p (x) is given. Find the zeroes of the polynomial.**

**Sol. **From the graph, it is evident that the graph of the given polynomial intersects the x-axis at - 1 and - 3.

∴ The zeroes of the polynomial are - 1 and - 3.**Q.17. If 1 is a zero of the polynomial p (x) = ax ^{2} - 3 (a - 1) x - 1, then find the value of a.**

∴ p(1) = a(1)

= a - 3a + 3 - 1

= - 2a + 2

∵ 1 is a zero of p (x);

∴ p (1) = 0

⇒ - 2a + 2 = 0 ⇒ a = 1

∵ (x + a) is a factor of p(x)

∴ - a is a zero of p(x)

⇒ p (- a) = 0

⇒ 2 (- a)

⇒ 2a

⇒ a = 10/5 = 2

Thus, the required value of a = 2.

Sol.

p (x) = x

= (x + 1)

If p (x) = 0 then (x + 1)

⇒ x = - 1

Thus, the zero of the polynomial p (x) is - 1.

p (x) = x

= x

⇒ x (x - 3) + 2 (x - 3)

⇒ (x - 3) (x + 2)

For p (x) = 0

⇒ (x - 3) (x + 2) = 0

∴ x = 3 and x = - 2

x

But, Sum of zeroes = 3

Product of zeroes = - 2

∴ The required quadratic polynomial

= x

= x

**Sol. **Since, curve (graph) of y = f (x) intersects the x-axis at three points,

∴ y = f (x) has three zeroes.**Q.23. The graph of y = f (x) is given in figure. How many zeroes are there of f (x)?**

**Sol.** The graph of y = f (x) does not intersect the x-axis.

∴ The number of zeroes of f (x) is 0.**Q.24. The graph of y = f (x) is given in the figure. What is the number of zeroes of f (x)?****Sol. **Since, the graph of y = f (x) intersects the x-axis at three distinct points,

∴ y = f (x) has 3 zeroes.**Q.25. What is the number of zeroes of the polynomial y = p (x)?****Sol. **Since, the graph passes through origin(0,0) which means on putting x = 0, we get y = 0.

Therefore, (0,0) is the only root of this equation.

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